Question: $\dfrac{d}{dx}\left(\dfrac{3x^2-1}{x-2}\right)=$
Answer: $\dfrac{3x^2-1}{x-2}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( 3 x 2 − 1 x − 2 ) = ( x − 2 ) d d x ( 3 x 2 − 1 ) − ( 3 x 2 − 1 ) d d x ( x − 2 ) ( x − 2 ) 2 = ( x − 2 ) ( 6 x ) − ( 3 x 2 − 1 ) ( 1 ) ( x − 2 ) 2 = 6 x 2 − 12 x − 3 x 2 + 1 ( x − 2 ) 2 = 3 x 2 − 12 x + 1 ( x − 2 ) 2 The quotient rule Differentiate ( 3 x 2 − 1 ) & ( x − 2 ) Expand \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{3x^2-1}{x-2}\right) \\\\ &=\dfrac{(x-2)\dfrac{d}{dx}(3x^2-1)-(3x^2-1)\dfrac{d}{dx}(x-2)}{(x-2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x-2)(6x)-(3x^2-1)(1)}{(x-2)^2}&&\gray{\text{Differentiate }(3x^2-1)\text{ & }(x-2)} \\\\ &=\dfrac{6x^2-12x-3x^2+1}{(x-2)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{3x^2-12x+1}{(x-2)^2} \end{aligned} In conclusion, $\dfrac{d}{dx}\left(\dfrac{3x^2-1}{x-2}\right)=\dfrac{3x^2-12x+1}{(x-2)^2}$, or any other equivalent form.